Problem 14.45 Two small spheres A and B, of mass 2.5 kg and 1 kg, respectively, are connected by a rigid rod of negligible weight. The two spheres are resting on a horizontal, frictionless surface when A is suddenly given the velocity Determine 0 v  (3.5 m/s) .i (a) the linear momentum of the system and its angular momentum about its mass center G, (b) the velocities of A and B after the rod AB has rotated through 180 .

The diagram showing the 2 spheres is missing, so i have attached it.

A) Linear momentum =(8.75 kg/m.s)i

Angular momentum =  (-0.5kg.m²/s)k

B) Va' = (1.5m/s)i and Vb' = (5 m/s)j

Explanation:

First of all, let's find the position of the mass centre;

y' = Σ(mi.yi)/mi = [2.5(0) + 1(0.2)]/(2.5+1) = 0.2/3.5 =0.057143 m

A) Linear momentum is given as;

L = m(a) x v(o) = (2.5 x 3.5)i = (8.75 kg/m.s)i

Angular momentum is given as;

HG = Vector GA  x m(a) x v(o)

Where vector GA is the position of the mass centre;

Thus;

HG = 0.057143j x (2.5 x 3.5)i = (-0.5 kg.m²/s) k

B) from conservation of linear momentum;

MaVo = Ma Va' + Mb Vb

2.5 x 3.5 = 2.5Va' + 1 Vb'

8.75 = 2.5Va' + 1 Vb' - - - - eq(1)

Also, for conservation of angular momentum ;

raMaVo = - raMa Va' + rbMb Vb'

from the diagram attached, ra + rb = 0.2

Now, ra is the same as  value as that of the centre of the mass.

Thus, ra = 0.057143.

rb = 0.2 - ra = 0.2 - 0.057143 = 0.14286

Thus;

0.057143 x 8.75 = -(0.057143 x 2.5)Va' + (0.14286 x 1) Vb'

0.5 = - 0.14286Va' + 0.14286Vb' - - - eq 2

Solving eq 1 and 2 simultaneously, we get; Va' = 1.5m/s and Vb' = 5 m/s

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